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RAVILLA DILLI, ECE, MIT, Manipal, India

 DIGITAL COMMUNICATIONS

Topics to be Discussed in this Section

Waveform Coding Techniques (Contd.,)

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 Digital Communications
Waveform Coding Techniques
Delta Modulation (DM):

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 Digital Communications
Waveform Coding Techniques
Delta Modulation (DM):

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 Digital Communications
Waveform Coding Techniques
Delta Modulation (DM):

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 Digital Communications
Waveform Coding Techniques
Delta Modulation (DM):

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 Digital Communications
Waveform Coding Techniques
Adaptive Delta Modulation (ADM):

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 Digital Communications
Waveform Coding Techniques
Adaptive Delta Modulation (ADM):

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 Digital Communications
Waveform Coding Techniques
Adaptive Delta Modulation (ADM):

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 Digital Communications
Waveform Coding Techniques
Adaptive Delta Modulation (ADM):

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 Digital Communications
Waveform Coding Techniques
Adaptive Delta Modulation (ADM):

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 Digital Communications
Waveform Coding Techniques
Adaptive Delta Modulation (ADM):

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 Digital Communications
Solved Examples
In a binary PCM, the no. of quantization levels, L = 2n or n = log2(L)
n  no. of bits to represent a quantization level.
If a signal m(t) is band-limited to B Hz, then we have to sample it at the rate
2B samples/sec.
It means the transmission bit rate is 2nB bps.
Therefore, the min. bandwidth BT = nB Hz

Let us consider a sampled signal m(t) and the quantized amplitude levels are
ranged from –mp to +mp.

Then the separation between the quantized levels, Δ = 2 mp/ L

L  No. of uniformly spaced intervals

We know that the mean square quantization error, σQ2 = Δ2/12

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 Digital Communications
Assignment-2
Problem 1: If the signal m(t) = 2cos(2π x 250 t) as the signal input. Find the
SNR with 8-bit PCM. Find the o/p SNR with 8-bit PCM.

Solution:
For 8-bit encoding, L= 256.
From the given signal, mp= 2 volts
The avg. signal power, Pavg = Am2/2 = 2 Watts
The uniform interval, Δ = 2 mp/L = 1.625 x 10-2 volt

The (SNR)O, Q = Pavg/ Quantization noise

= Pavg/(Δ2/12)

= 98,304

= 49.93 dB
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 Digital Communications
Assignment-2
Problem 2: If the signal m(t) = 2cos(2π x 250 t) as the signal input. If the
minimum SNR is to be at least 36 dB, how many bits are needed to encode
the signal level?
Solution: 36 dB equivalent is 3981
Δ = 2 mp/L = 4/L

The (SNR)O, Q = [mp2/2]/ [Δ2/12]

3981 = 2/[16/L2] x 12

3981 = 1.5 L2  L = 51.51

Therefore, the min. no. of bits required to represent a signal level = 6-bits

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 Digital Communications
Solved Examples
Problem 3: Consider a band-limited signal m(t) of 3 kHz bandwidth is
sampled at a rate 33.33% higher than the Nyquist’s rate. The maximum
allowable quantization error is 0.5% of the peak amplitude mp. Find the min.
bandwidth of the channel to transmit the binary encoded signal.

Solution: The Nyquist’s rate = 2 x 3kHz = 6kHz (samples/sec)

The given sampling rate is 33.33% higher than Nyquist’s rate.
= 6kHz + (6000/3) = 8kHz

We know that the separation between the quantized levels, Δ = 2 mp/ L

L  No. of uniformly spaced intervals
From the given data, Δ/2 = 0.005 mp  Δ = 0.01 mp

Therefore, 0.01 mp = 2mp/L  L = 200

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 Digital Communications
Solved Examples
For binary coding, L, must be a powers of two;
Therefore, L = 27 = 128 and 28 = 256,
we must choose n = 8 to guarantee better than a 0.5% error.

The transmission rate, C= 8kHz x 8-bits = 64 kbps

The transmission bandwidth, BT= C/2 = 32 kHz

NOTE: 2-bits/sec can be transmitted over 1 Hz of bandwidth.

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 Digital Communications
Solved Examples
Problem 4: A band-limited signal of bandwidth 10 MHz is binary encoded
using PCM.
a) What is the Nyquist’s rate?

b) If the samples are encoded into 128 level, how many no. of bits are
required to represent each sample?

c) What is the min. bitrate of the binary encoded signal?

d) What is the min. bandwidth required to successfully transmit this signal?

Solution: a) 20 MHz b) 7-bits c) 140 Mbps d) 70 MHz

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 Digital Communications
Solved Examples
Problem 5: A band-limited signal whose highest freq. component is 4.2 MHz
and 4 volts(p-p) is transmitted using a binary PCM. If the quantization levels
are 512 and normalized signal power is 0.04 Watts, find:
a) code word length b) bit rate c) o/p signal to quantization noise ratio.

Solution: a) 9-bits
b) 2fm samples/sec= 2 x 4.2 MHz x 9-bits = 75.6 Mbps
c) Normalized signal power/Normalized Quantization noise power
= 0.04 / (Δ2/12) = 7864.32 = 38.96 dB

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 Digital Communications
Solved Examples
Problem 6: If the signal x(t) is modeled as the sample function of zero mean
stationary random process X(t) with uniform pdf in the range (-a, a). Find the
(SNR)O, Q assuming an R-bit code word/ sample.

Solution: As X(t) is uniformly distributed over the range (-a, a),

its pdf is given as, fX(x) = 1/2a, -a < x < a

Variance, σX2 = E[X2] – E[X]2 = a2/3

Step size, Δ = 2a/L = 2a/2R

The mean square quantization error, σQ2 = Δ2/12 = 22R

(SNR)O, Q in dB = 6.02 R

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 Digital Communications
Solved Examples
Problem 7: If a sinusoidal signal x(t) with the peak value of ‘A’ is applied to
an uniform quantizer, find the o/p signal to quantization noise ratio assuming
R-bit code word per sample.

Solution: The signal power = A2/2

Step size, Δ = 2A/L = 2A/2R

(SNR)O, Q = [A2/2]/[Δ2/12] = 1.5 x (22R)

(SNR)O, Q in dB = 6.02 R + 1.8

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 Digital Communications
Solved Examples
In general, the (SNR)O, Q of a PCM system using uniform quantizer can be
taken as 6R + α, R  No. of bits/code word
α  a constant
and depends on the ratio [ Xrms/Xmax] = [σX /Xmax]
Solution: The signal power = A2/2
Step size, Δ = 2Xmax/L = 2Xmax/2R

(SNR)O, Q = σX2/ σQ2

= [σX2]/[Δ2/12]
= 3 [σX /Xmax]2 22R

(SNR)O, Q in dB =4.77 + 6.02 R + 20 log10 [σX /Xmax]

= 6.02R + α
where, α = 4.77 + 20 log10 [σX /Xmax]

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 Digital Communications
Solved Examples
If a real band-pass signal m(t) with a non-zero spectrum (for f > 0) only in the
range fL to fH, (fH ≥ fL),

then the min. sampling freq. to avoid aliasing is (fs)min = 2fH/ ꓡk˩

where k = fH/B, B = fH - fL

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 Digital Communications
Solved Examples
Problem 8: Let a band-pass signal m(t) with fL= 3 MHz and fH = 5 MHz is sent
using PCM on a channel whose capacity is limited to 7 Mbps and the samples
are uniformly distributed in the range -2 V to + 2 V. Find:
a) the min. sampling rate to avoid aliasing
b) The no. of uniform quantization levels
c) (SNR)O, Q in dB.
Solution:
a) the min. sampling freq. to avoid aliasing is (fs)min = 2fH/ ꓡk˩

where k = fH/B, B = fH – fL

= 5 x 106 samples/sec

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 Digital Communications
Solved Examples
Solution:
b) The sampling rate = 5 x 106 samples/sec
Capacity of the channel is limited to 7Mbps;

Therefore, it is possible to send only one bit per sample;

i.e., we have only a two-level quantization.
The quantizer levels are at ±1 so that step size ∆ is 2 V
and the error is uniform in the range ±1.

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 Digital Communications
Solved Examples
Solution:
c) As the signal is uniformly distributed over (-2, 2),
its pdf, fM(m) = 1/4, -2 < m < 2
= 0, otherwise

σM2 = 4/3 σQ2 = 1/3

(SNR)O, Q = 4, (SNR)O, Q in dB = 6.02 dB

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 Digital Communications
Solved Examples
Problem 9: Consider the characteristics of the quantizer as shown below.
Let X is the i/p to the quantizer with pdf fX(x). Find:
a) value of ‘A’
b) Total quantization noise variance
c) is it same as Δ2/12 ?

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 Digital Communications
Solved Examples
Solution:
a) A = 1/4 (since the area under the curve is unity)

b) fX(x) = 1/4 -(1/16)|x|, |x|≤ 4

= 0, otherwise

Total variance of quantization noise = 2 x ( quantization noise for x ≥ 0)

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 Digital Communications
Solved Examples
Problem 10: If the i/p to the quantizer shown in problem 6 is a random
Variable with the following pdf, find:
a) value of ‘A’
b) Total quantization noise variance
c) is it same as Δ2/12 ?
Solution:

Problem 11: If a uniformly distributed random variable over 0 to 1 is

quantized as follows:
QZ(x) = 0, 0 ≤ x ≤ 0.3
= 0.7, 0.3 ≤ x ≤ 1
Show that the rms value of the quantization noise is 0.198.

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 DIGITAL COMMUNICATIONS

Topics to be Discussed in this Section

Baseband Shaping for Data Transmission

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 QUERIES ?

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17-Mar-19 Dr. RAVILLA DILLI, ECE, MIT, Manipal, India